Talk:Period-50 glider gun

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Discoverer?

I'm reading over some of the old archives, and it looks to me like this gun was discovered on Oct. 13, 1996 by Dean Hickerson, not David Buckingham (as listed on the page right now). Admittedly it used Buckingham's technology rather directly, but in my mind Hickerson is still the "discoverer" of this pattern. Thoughts? Nathaniel (talk) 17:29, 20 February 2017 (UTC)

I agree, and so does Jason Summers. I've have made the corrections, although we should really add multiple discoverers for the infoboxes, since Buckingham's (and I think Elkies') contributions were significant.
~Sokwe 20:22, 22 February 2017 (UTC)

Smaller true-period gun

It is possible to shrink the true-period p50 gun slightly by using different p5 sparkers:

x = 112, y = 38, rule = B3/S23
32b2o2b2o36b2o2b2o$32bo2bob3o32b3obo2bo$33b3o4bo10b2o6b2o10bo4b3o$26bo
b2o7b2obo10b2o6b2o10bob2o7b2obo$26b2o2bo4b2obob2o28b2obob2o4bo2b2o$29b
2o3bo4bo2bo26bo2bo4bo3b2o$6bo2bo16b2obo11bo28bo11bob2o16bo2bo$6bo2bo
16b2obob7ob2ob3o22b3ob2ob7obob2o16bo2bo$4b2ob2ob2o18b3obo7bo2bo20bo2bo
7bob3o18b2ob2ob2o$6bo2bo24b2ob4o3bo7bo6bo7bo3b4ob2o24bo2bo$6bo2bo26bo
3b4o7b3o4b3o7b4o3bo26bo2bo$4b2ob2ob2o38bo2b2o2b2o2bo38b2ob2ob2o$6bo2bo
22bo9b2o5b4obo2bob4o5b2o9bo22bo2bo$6bo2bo16bo15b2o6b5o2b5o6b2o15bo16bo
2bo$12b3o11b4o5bo17b2o2b2o17bo5b4o11b3o$2o23bob2o54b2obo23b2o$o2b2o5bo
5bo5b2o5bo5b3o9b2o14b2o9b3o5bo5b2o5bo5bo5b2o2bo$b2ob2o4bo5bo5b2obob2ob
4obobo8b3ob3o6b3ob3o8bobob4ob2obob2o5bo5bo4b2ob2o$4b2o4bo5bo5bobob4ob
2obob2o9b2o14b2o9b2obob2ob4obobo5bo5bo4b2o$4b2o16b3o5bo5b2o36b2o5bo5b
3o16b2o$b2ob2o6b3o16b2obo18b2o2b2o18bob2o16b3o6b2ob2o$o2b2o19bo5b4o8b
2o6b5o2b5o6b2o8b4o5bo19b2o2bo$2o31bo8b2o6b3obo2bob4o5b2o8bo31b2o$27bo
19b4o2b2o2b2o2bo22bo$47bobob3o4b3o7b4o3bo$32bob6o3b2o3bo3bo6bo7bo3b4ob
2o$13b2o3bo11bo2bob3o2bo2bo2bobo17bo2bo7bob3o$13bobo2b3o12bo5bo5b2o2b
2o16b3ob2ob7obob2o$15b3o3bo6b4o14bo3bo19bo11bob2o$14bo3bobobob2o2bobo
6b2o5bo4bo19bo2bo4bo3b2o$13bob2obobo3bob2o3bo2b2ob2o5b2o3b2o4bo14b2obo
b2o4bo2b2o$13bo4bobo2b3o2bobobo2b3o16bobo14bob2o7b2obo$11b2obo3b2obo2b
o2b2o2bo3b3o17bo15bo4b3o$10bo3bobo3bobo2bobo5bo3bo34b3obo2bo$11b3o4b3o
b4o2bo5bobo37b2o2b2o$14b4o3bo4b2o5b2obobo31bo$13bo3bo2bo2b2o2bobo7b2o
32bo$13b2o6b2ob2o2b2o39b3o!

There are probably even smaller solutions.
~Sokwe 20:56, 22 February 2017 (UTC)